Array of pointer initialization

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September 4th, 2012, 12:43 AM

mahaju

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Array of pointer initialization

Code:

    Fcomplex enhance_arg[half_FFT] = {{1.0,1.1},
                                      {2.0,2.1},
                                      {3.0,3.1},
                                      {4.0,4.1},
                                      {5.0,5.1},
                                      {6.0,6.1},
                                      {7.0,7.1},
                                      {8.0,8.1},
                                      {9.0,9.1},
                                      {10.0,10.1} };

    Fcomplex ubc_arg[OCTO_MAX_MIC_NUMBER-1][half_FFT] = {
                                                        { {1.0, 1.1},{2.0, 2.1},{3.0, 3.1},{4.0, 4.1},{5.0, 5.1},{6.0, 6.1},{7.0, 7.1},{8.0, 8.1},{9.0, 9.1},{10.0, 10.1}},
                                                        { {11.0, 11.1},{22.0, 22.1},{33.0, 33.1},{44.0, 44.1},{55.0, 55.1},{66.0, 66.1},{77.0, 77.1},{88.0, 88.1},{99.0, 99.1},{100.0, 100.1}},
                                                        { {111.0, 111.1},{222.0, 222.1},{333.0, 333.1},{444.0, 444.1},{555.0, 555.1},{666.0, 666.1},{777.0, 777.1},{888.0, 888.1},{999.0, 999.1},{1000.0, 1000.1}},
                                                        { {1111.0, 1111.1},{2222.0, 2222.1},{3333.0, 3333.1},{4444.0, 4444.1},{5555.0, 5555.1},{6666.0, 6666.1},{7777.0, 7777.1},{8888.0, 8888.1},{9999.0, 9999.1},{10000.0, 10000.1}}
                                                        };

typedef struct FCOMPLEX
{
	float r,i;
} Fcomplex;

In this declaration, if i go into single step mode and view the value of ubc_arg and enhance_arg in the watch window, I can see that ubc_arg[0][0].r = 1 and ubc_arg[0][0].i = 1.10000002 while enhance_arg[5].r = 6 and enhance_arg[5].i = 6.0999999999

This is true for any array index and not just these particular ones

Why don't they have exact values in the watch window?

Using gcc version 4.4.1 (TDM-2 mingw32), in CodeBlocks 10.05

September 4th, 2012, 12:59 AM

Scorpions4ever

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This is actually a problem of representing floating point numbers and exists for all computers, irrespective of the programming language used. Remember that the computer stores everything in binary internally and certain numbers that are easily expressed in one base (e.g. decimal) cannot be exactly represented in another base (e.g. binary).

See http://c-faq.com/fp/printfprec.html for more details.

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September 4th, 2012, 01:34 AM

mahaju

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I can understand why it is happening
but I thought it took place only when we had to work on large precision numbers such as those with 8 or 10 digits after the decimal point
I thought 1.1 was a fairly large number so would be displayed exactly
So, would it be possible to list out (say, for eg, for numbers with 2 or 3 digits after decimal) those numbers which can be exactly represented?
Can this is done using C code itself, or is there a table I can look up if I want?

September 4th, 2012, 02:13 AM

b49P23TIvg

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How, dude or dudette, please, would you exactly represent 1/10 (one tenth) (0.1) exactly in base 2 ?

How would you represent 2/3 exactly in base 10?

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September 7th, 2012, 01:32 AM

Lux Perpetua

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Quote:

Originally Posted by mahaju

The only numbers that can be exactly represented as a float or double are of the form N/2^m, where N and m are integers. For example, 5/16 can be represented exactly because 16 is a power of two. However, using an example from your post, 1.1 = 11/10 cannot be written in the form N/2^m, so it cannot be represented exactly.

September 7th, 2012, 01:40 AM

mahaju

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Quote:

Originally Posted by Lux Perpetua

Thank you very much
That was exactly what I was looking for

Another question, is there a way I can view the exact contents of the memory register containing a float number? For example, if I have a variable float a = 5.1, how can I see the exact sequence of 0 and 1 in the 4 bytes of memory represented by 'a'? is there a way to printf the exact bit patterns to the screen?

September 7th, 2012, 03:22 AM

Lux Perpetua

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Something like this?

c Code:

Original - c Code

#include <stdio.h>
#include <stdlib.h>
 
/* Comment this for big-endian byte order */
#define LITTLE_ENDIAN
 
int binary_char(char *s, unsigned char c) {
    int k;
    for (k = 7; k >= 0; --k)
        *(s++) = "01"[(c>>k)&0x1];
    *s = '\0';
 
    return 8;
}
 
void print_bits(void *p, int n) {
    char buf[9];
    int k;
 
#ifdef LITTLE_ENDIAN
    for (k = n - 1; k >= 0; --k)
#else
    for (k = 0; k < n; ++k)
#endif
    {
        unsigned char c = ((unsigned char *)p)[k];
        binary_char(buf, c);
        printf("%s ", buf);
    }
}
 
int main(int argc, char **argv) {
    double x;
    float xf;
 
    if (argc < 2) {
        fprintf(stderr, "Usage: %s <float>\n",
                argv[0]);
        exit(2);
    }
 
    x = atof(argv[1]);
    xf = (float)x;
 
    printf("double: ");
    print_bits(&x, sizeof x);
    printf("\nfloat: ");
    print_bits(&xf, sizeof xf);
    printf("\n");
 
    return 0;
}

To interpret the output, this might be helpful: http://en.wikipedia.org/wiki/Floati...odern_computers

Array of pointer initialization

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